If it's not what You are looking for type in the equation solver your own equation and let us solve it.
-4=-q2
We move all terms to the left:
-4-(-q2)=0
We add all the numbers together, and all the variables
-(-1q^2)-4=0
We get rid of parentheses
1q^2-4=0
We add all the numbers together, and all the variables
q^2-4=0
a = 1; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·1·(-4)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*1}=\frac{-4}{2} =-2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*1}=\frac{4}{2} =2 $
| 4(x-1.9)=5.2 | | 5s2+8s=-4 | | 5q+15=3q+1 | | 8(10-k=2k | | (4x+1)=28 | | 7r2+3r–4=0 | | (8-6i)+(7+4i)=0 | | 5(3v-5)(v+4)=0 | | 9w2+8=5w | | 5(x+3)=3(x+5 | | 4p+25=-p^2+30p | | 3t2=4t | | 2/3=(4u-5) | | 5y2–4y–6=0 | | 4/5(5x+1)=28 | | 3b+12=3(b−5) | | 4r2+4r+1=0 | | x2-32=0 | | z2+7z+9=0 | | 9p+5=-3 | | 4÷5(4x+1)=28 | | 2z2=-5z | | 2(2a+10)=5(a+5) | | 7x+7x+5x=180 | | (x-3)4=24 | | 2p2+9=0 | | x+(x+12)+2(x+12)=216 | | 4r2+1=0 | | -1/4x+15=-20 | | 3v2–5=0 | | 2x×5=4 | | 9x-23=1 |